Lesson In preparation for Halloween
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<H2>In preparation for Halloween</H2> <H3>Problem 1</H3>In preparation for Halloween, three married couples, the Browns, the Joneses and the Smiths, bought little presents for the neighborhood children. Each bought as many identical presents as he (she) paid cents for one of them. Each wife paid 75 cents more the her husband, Ann bought one more present than Bill Brown, Betty one less present than Joe Jones. What is Mary's last name? <B>Solution</B> <pre> 1. From the condition, it is clear that the amounts each person spent for presents, each (amount) is a perfect square of cents. 2. If x and y are spendings for some (any) of the three couples, then x^2 - y^2 = 75, according to the condition, or (x+y)(x-y) = 75. (and we remember that the greater value is the wife's spending !) For integer positive x and y it gives these and only these opportunities x + y = 75 x - y = 1 with the solution x= 38, y= 37 OR x + y = 25 x - y = 3 with the solution x= 14, y= 11 OR x + y = 15 x - y = 5 with the solution x= 10, y= 5. Of these solutions, the only pair (38,37) has the difference of 1, so it gives a clue to me to conclude that Ann is the wife to Bill Brown. 3. But in order for to make this conclusion ABSOLUTELY CORRECT, I must <U>EXCLUDE</U> that the other couples fall in the same pair/solution (38,37). Fortunately, the condition gives me rationale to make this conclusion. Indeed, it says that "Betty bought one less present than Joe Jones", which means that Betty is of the pair (10,5), while Joe Jones is of the pair (14,11). So, I really can conclude that Ann is the wife to Bill Brown. 4. Finally, from the condition, it is also clear that Betty <U>IS NOT</U> the wife to Joe Jones. 5. 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